The process of calculating the reaction in a beam<\/h1>\n\n\n\n\n- We start by introducing the appropriate support reactions at the support site. Read more about this in the publication Reaction forces<\/a>.<\/li>\n\n\n\n
- Then we check whether the ray is statically determinate. Read more about this in the publication statically determinate<\/a>.<\/li>\n\n\n\n
- In the next step, we write down the equilibrium equations. Read more about this in the publication Equilibrium equations<\/a>.<\/li>\n<\/ul>\n\n\n\n
Freely supported beam-calculation of support reactions for beams<\/h2>\n\n\n\n
We start by taking a coordinate system and assuming a positive counterclockwise torque.<\/p>\n\n\n\n![\"Marking](\"https:\/\/i0.wp.com\/solveredu.com\/wp-content\/uploads\/2024\/09\/uklad-wspol.png?resize=629%2C591&ssl=1\")
<\/figure>\n\n\n\nBelow I have included an example of a freely supported beam diagram. This is what we call a beam, supported by articulated supports at both ends. We will assign support reactions for this beam.<\/p>\n\n\n\n![\"freely](\"https:\/\/i0.wp.com\/solveredu.com\/wp-content\/uploads\/2024\/09\/belka_swobodna.png?resize=1024%2C340&ssl=1\")
<\/figure>\n\n\n\nReactions have already been added to the beams drawing. So, at point A, we have a pivot bearing support, so we add a horizontal reaction HA and a vertical va. At point B at the end of the beam, we have a hinged sliding support, so we add one vertical reaction VB. Then let's check the static notability.<\/p>\n\n\n\nN=R-J-3<\/em>=3-0-3=0 - the beam is statically determinable
where:
N - degree of static indeterminacy
R =3 - number of support reactions
J =0 - number of internal joints
3 - the number of equilibrium equations. In static systems it is 3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\nIt's time for the equilibrium equations. Remember that for a planar system of forces, we have three equations:<\/p>\n\n\n\n
![\"\\F_{ix}](\"https:\/\/s0.wp.com\/latex.php?latex=%5CSigma+F_%7Bix%7D+%3D+0+&bg=ffffff&fg=000&s=2&c=20201002\")
- sum of the projections of forces on the x-axis<\/p>\n\n\n\n
![\"\\F_{iy}](\"https:\/\/s0.wp.com\/latex.php?latex=%5CSigma+F_%7Biy%7D+%3D+0+&bg=ffffff&fg=000&s=2&c=20201002\")
- sum of the projections of forces on the y-axis<\/p>\n\n\n\n
![\"\\\u25a0Sigma](\"https:\/\/s0.wp.com\/latex.php?latex=%5CSigma+M_%7Bi%7D+%3D+0+&bg=ffffff&fg=000&s=2&c=20201002\")
- sum of moments at a point<\/p>\n\n\n\n
Let's start with the first and simplest equation. Sum of force projections on the x-axis.<\/p>\n\n\n\n![\"Equation](\"https:\/\/i0.wp.com\/solveredu.com\/wp-content\/uploads\/2024\/09\/swobodna-Fx.png?resize=271%2C51&ssl=1\")
<\/figure>\n\n\n\nSince in our example a freely supported beam does not have any component of the force acting in the x-axis direction, the reaction HA=0.<\/p>\n\n\n\n
Then we move on to the third equation, the sum of moments at a point.<\/p>\n\n\n\n
The choice of point is up to you. I chose point A.<\/p>\n\n\n\nWhen determining the equilibrium equations for the sum of moments, it is best to choose the point at which one of the supports is located.<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\nIn our example, we have the choice of punk A or B. By choosing one of the supports, we make the reaction for that support not appear in our equation for the moment, because the moment is the force multiplied by the hand. If the arm is zero (the force passes through our point a), then the moment of this force will also be zero, so we can omit it in the equation.<\/p>\n\n\n\n![\"Bending](\"https:\/\/i0.wp.com\/solveredu.com\/wp-content\/uploads\/2024\/09\/swobodna-moment.png?resize=634%2C108&ssl=1\")
<\/figure>\n\n\n\nIn the equation, we have:<\/p>\n\n\n\n
\n- Reaction VB multiplied by the distance 12, that is, as much as between points A and B.<\/li>\n\n\n\n
- Force F multiplied by 2, i.e. the distance of force F from point A<\/li>\n\n\n\n
- We do not multiply the bending moment M of the moment by the distance.<\/li>\n\n\n\n
- The continuous load q is multiplied by the length 4 on which it acts and 6, that is, the distance of the center of q to point A.<\/li>\n\n\n\n
- We pay attention to the signs of moments in accordance with what we assumed at the beginning in Fig. 1<\/li>\n<\/ul>\n\n\n\n
After the transformations, we get the force value VB, so we get the calculated reaction.<\/p>\n\n\n\n
Finally, we will write an equilibrium equation for forces towards the y-axis.<\/p>\n\n\n\n![\"Equation](\"https:\/\/i0.wp.com\/solveredu.com\/wp-content\/uploads\/2024\/09\/swobodna-Fy.png?resize=483%2C113&ssl=1\")
<\/figure>\n\n\n\nIn the equation, we have:<\/p>\n\n\n\n
\n- The VA response is positive, since the return of the VA force corresponds to the return of the y-axis<\/li>\n\n\n\n
- Reaction VB with a positive sign, since the return of force VA corresponds to the return of the y-axis<\/li>\n\n\n\n
- Continuous load q multiplied by 4, i.e. the length at which it operates<\/li>\n\n\n\n
- Force F with the sign ujemy, because the return of force F is opposite to the y-axis<\/li>\n<\/ul>\n\n\n\n
After converting and substituting VB values, we get the strength value VA. So we calculated all the reactions.<\/p>\n\n\n\n
Freely supported beam-calculation of support reactions for beams<\/h2>\n\n\n\n
We start by taking a coordinate system and assuming a positive counterclockwise torque.<\/p>\n\n\n\n Below I have included an example of a freely supported beam diagram. This is what we call a beam, supported by articulated supports at both ends. We will assign support reactions for this beam.<\/p>\n\n\n\n Reactions have already been added to the beams drawing. So, at point A, we have a pivot bearing support, so we add a horizontal reaction HA and a vertical va. At point B at the end of the beam, we have a hinged sliding support, so we add one vertical reaction VB. Then let's check the static notability.<\/p>\n\n\n\n It's time for the equilibrium equations. Remember that for a planar system of forces, we have three equations:<\/p>\n\n\n\n Let's start with the first and simplest equation. Sum of force projections on the x-axis.<\/p>\n\n\n\n Since in our example a freely supported beam does not have any component of the force acting in the x-axis direction, the reaction HA=0.<\/p>\n\n\n\n Then we move on to the third equation, the sum of moments at a point.<\/p>\n\n\n\n The choice of point is up to you. I chose point A.<\/p>\n\n\n\n In our example, we have the choice of punk A or B. By choosing one of the supports, we make the reaction for that support not appear in our equation for the moment, because the moment is the force multiplied by the hand. If the arm is zero (the force passes through our point a), then the moment of this force will also be zero, so we can omit it in the equation.<\/p>\n\n\n\n In the equation, we have:<\/p>\n\n\n\n After the transformations, we get the force value VB, so we get the calculated reaction.<\/p>\n\n\n\n Finally, we will write an equilibrium equation for forces towards the y-axis.<\/p>\n\n\n\n In the equation, we have:<\/p>\n\n\n\n After converting and substituting VB values, we get the strength value VA. So we calculated all the reactions.<\/p>\n\n\n\n<\/figure>\n\n\n\n<\/figure>\n\n\n\nN=R-J-3<\/em>=3-0-3=0 - the beam is statically determinable
where:
N - degree of static indeterminacy
R =3 - number of support reactions
J =0 - number of internal joints
3 - the number of equilibrium equations. In static systems it is 3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<\/figure>\n\n\n\nWhen determining the equilibrium equations for the sum of moments, it is best to choose the point at which one of the supports is located.<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n <\/figure>\n\n\n\n\n
<\/figure>\n\n\n\n\n
![\"The](\"https:\/\/i0.wp.com\/solveredu.com\/wp-content\/uploads\/2024\/09\/swobodna-calosc-1.png?resize=639%2C299&ssl=1\")
<\/figure>\n\n\n\n![\"Cantilever](\"https:\/\/i0.wp.com\/solveredu.com\/wp-content\/uploads\/2024\/09\/belka_wspornik.png?resize=900%2C244&ssl=1\")
<\/figure>\n\n\n\n![\"Equations](\"https:\/\/i0.wp.com\/solveredu.com\/wp-content\/uploads\/2024\/09\/wspornik_fx.png?resize=458%2C118&ssl=1\")
<\/figure>\n\n\n\n![\"Equation](\"https:\/\/i0.wp.com\/solveredu.com\/wp-content\/uploads\/2024\/09\/wspornik-fy.png?resize=496%2C109&ssl=1\")
<\/figure>\n\n\n\n![\"Equation](\"https:\/\/i0.wp.com\/solveredu.com\/wp-content\/uploads\/2024\/09\/wspornik_Ma.png?resize=685%2C125&ssl=1\")
<\/figure>\n\n\n\n![\"The](\"https:\/\/i0.wp.com\/solveredu.com\/wp-content\/uploads\/2024\/09\/calosc_wspornik.png?resize=702%2C367&ssl=1\")
<\/figure>\n\n\n\n